The wave equation in $3$ dimensions is simply:

$$\nabla^2\psi = \dfrac{1}{v^2} \dfrac{\partial^2}{\partial t^2}\psi,$$

and the intuition behind this is that at each point of space with coordinates $(x,y,z)$ we have some quantity oscillating there. If it's a sound wave what is oscillating are molecules, if it's an electromagnetic wave what is oscillating are electromagnetic fields and so on. The important thing is: $\psi$ represents the association $t\mapsto \psi_t$ where $\psi_t$ represents a quantity on each point of space and this association from $t$ into $\psi_t$ is to be thought of "oscillatory".

Basic physics texts take the oscillation of points on a string, derive from Newton's second law that the wave equation in $1$-dimension is obeyed and then say: "because of that we have good reasons already to call wave something that satisfies this equation".

The problem is that I'm not yet convinced. Is there any other way to "derive" the wave equation without referring to the particular case of waves on strings? That is, starting from the fact that we want that association $t\to\psi_t$ as I've said, is there a way to conclude that the $\psi$ function should satisfy that equation?

I've tried to reason with the harmonic oscillator. So at each point $(x_0,y_0,z_0)$ the function $t\mapsto \psi(t,x_0,y_0,z_0)$ should satisfy the harmonic oscillator equation, that is:

$$\dfrac{\partial^2}{\partial t^2}\psi + \omega^2 \psi =0,$$

But I think that's not the way, since I can't see a way to put the Laplacian in there. So how can we do it?

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    $\begingroup$ Optics by Hetch make a pure mathematical derivation. It might interest you. $\endgroup$
    – jinawee
    Apr 7, 2014 at 21:57
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    $\begingroup$ I guess it depends on what you mean by "derive". Typically in physics we don't say "can we find a differential equation that is satisfied by this function $f(\mathbf{r},t)?$", but rather we say "we have a system with these properties; what differential equation governs its behavior?". $\endgroup$ Apr 7, 2014 at 23:08
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    $\begingroup$ That's the point @DumpsterDoofus, I just think I've missed the right word to explain. What I'm really questioning is why that differential equation governs wave motion? How could we realize that's the equation without resorting to a particular case, but rather to the properties of the phenomenon it should describe. $\endgroup$
    – Gold
    Apr 7, 2014 at 23:26
  • $\begingroup$ I still don't quite see what you are after. ( so I'm with @DumpsterDoofus for now. ) I don't understand the phrase "but rather to the properties of the phenomenon it should describe" $\endgroup$
    – garyp
    Apr 8, 2014 at 1:25
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    $\begingroup$ I'll try to explain again. What is a wave intuitively? By what I've learned, it's a way to transport energy without transporting matter and that has oscilatory behavior. This means that a wave should be described by a map $\psi : \mathbb{R}^4\to \mathbb{R}$ with the property that $\psi(t,x_0,y_0,z_0)$ oscilates when $t$ varies. From these assumptions, is it possible to conclude that whatever $\psi$ is it must satisfy the wave equation? $\endgroup$
    – Gold
    Apr 8, 2014 at 1:29

3 Answers 3


I think the wave equation can by derived from geometry alone, without using physics. Consider $f(x-ct)$ and consider small changes in $x$ and $t$, ie. $\Delta x$, $\Delta t$ (They each cause a small shift or translation of $f(x-ct)$). Note that $\Delta x$ = $c\Delta t$. So $\frac{\Delta f}{\Delta x}$ = $\frac{\Delta f}{c\Delta t}$ = $\frac{1}{c}\frac{\Delta f}{\Delta t}$. Doing that again we get $$\frac{\Delta^2 f}{\Delta^2 x} = \left(\frac{1}{c}\right)^2\frac{\Delta^2 f}{\Delta^2 t}$$. Then letting $\Delta$ become very small we get

$$ \frac{\partial^2f}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2f}{\partial t^2}=0 $$

From the geometry alone, it was only needed to note that a change in $t$ multiplied by the velocity yields the same results (as measured by the second derivative) as a change in $x$ --that is, a translation of $f(x-ct)$. See, for example, for a good discussion: kiskis.physics.ucdavis.edu/landau/phy9hc_03/wave.pdf‎.

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    $\begingroup$ More or less the answer I was about to write :). It can be made easier perhaps by looking at the functions f(x-ct) and f(x+ct) and just taking the second derivatives to space and time ? $\endgroup$
    – Nick
    May 3, 2014 at 17:38
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    $\begingroup$ That's a very good short derivation too. (Landau did that in the above pdf). It also shows that physics is not needed in the wave equation derivation. I think that approach is slightly more mathematical and less geometric though, and maybe geometry is a little more basic. $\endgroup$
    – user45664
    May 3, 2014 at 18:48
  • $\begingroup$ See physics.stackexchange.com/a/395212/45664 for an expanded version of this answer. $\endgroup$
    – user45664
    Mar 26, 2018 at 20:02
  • $\begingroup$ The link in the answer has changed to: kiskis.physics.ucdavis.edu/student/phy9b_11/wave.pdf $\endgroup$
    – user45664
    May 7, 2018 at 3:45
  • $\begingroup$ See physics.stackexchange.com/a/403761/45664 where the wave equation is mathematically derived directly from f(x-ct) alone. $\endgroup$
    – user45664
    May 22, 2018 at 3:00

The fact that the wave equation is ubiquitous in physics does not mean that the derivation of it is the same for each physical situation. I'll show you how to derive the wave equation of electrodynamics since it is pretty elegant and point you to some places to look at the derivation for other physical situations. For electromagnetic waves in vacuum, start with Maxwell's equations $$ \begin{align} \nabla\cdot\vec{E}&=0\qquad\qquad\nabla\times\vec{E}=-\frac{1}{c}\frac{\partial\vec{B}}{\partial t}\\ \nabla\cdot\vec{B}&=0\qquad\qquad\nabla\times\vec{B}=\frac{1}{c}\frac{\partial\vec{E}}{\partial t}. \end{align} $$ Take the curl of the the Faraday equation (upper right) and apply some vector identities $$ \begin{align} \nabla\times(\nabla\times \vec{E})&=-\frac{1}{c}\nabla\times\frac{\partial\vec{B}}{\partial t}\\ \nabla(\nabla\cdot\vec{E})-\nabla^2\vec{E}&=-\frac{1}{c}\frac{\partial}{\partial t}(\nabla\times\vec{B})\\ \nabla^2\vec{E}&=\frac{1}{c^2}\frac{\partial^2\vec{E}}{\partial t^2}. \end{align} $$ You can get a similar result for the magnetic field by taking the curl of Ampere's law. The derivation with a source is slightly more complicated. It is usually derived in terms of the vector, $\vec{A}$, and scalar, $\Phi$ potentials defined by $$ \vec{B}=\nabla\times\vec{A}\qquad\qquad\vec{E}=-\nabla\Phi-\frac{\partial\vec{A}}{\partial t}. $$ Maxwell's equations can be written in terms of these potentials as $$ \begin{align} \nabla^2\Phi+\frac{\partial}{\partial t}\nabla\cdot\vec{A}&=-\frac{\rho}{\epsilon}\\ \nabla^2\vec{A}-\frac{1}{c^2}\frac{\partial^2\vec{A}}{\partial t^2} -\nabla\left(\nabla\cdot\vec{A}+\frac{1}{c^2}\frac{\partial\Phi}{\partial t}\right)&= -\mu_0\vec{J}, \end{align} $$ where $\rho$ is the charge density and $\vec{J}$ is the current density, i.e. the sources. There is some freedom in defining these potentials known as gauge freedom. Gauge freedom is sacred to theroetical phycics, but I won't go into detail here. One choice of the freedom is to demand that the potentials satisfy what is known as the Lorenz condition $$ \nabla\cdot\vec{A}+\frac{1}{c^2}\frac{\partial\Phi}{\partial t}=0. $$ Plugging this into the above equations shows that both potentials satisfy the sourced wave equations $$ \begin{align} \nabla^2\Phi-\frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2}&=-\frac{\rho}{\epsilon_0}\\ \nabla^2\vec{A}-\frac{1}{c^2}\frac{\partial^2\vec{A}}{\partial t^2}&= -\mu_0\vec{J}. \end{align} $$ For more details on the electromagnetic wave equation have a look at chapter 6 of John David Jackson's classic text "Classical Electrodynamics".

For a derivation of the wave equation on a stretched membrane, like a drum head, have a look at these notes. For a derivation of the acoustic wave equation, have a look at the Wikipedia article.


I think you can make a rough heuristic like this. What we call wave motion involves some amplitude, and the wave propagates when the amplitude varies in space. The Laplacian operator has the mean-value property that when $\nabla^2 f = 0$ in a region $U$ and $x \in U$, $$f(x) = \frac{1}{V}\int_B f(t)\, dV$$ where $B$ is some ball centered around $x$ and contained in $U$, and $V$ is the volume of $B$. Thus we can say that the Laplacian measures how different $f(x)$ is from the mean value of $f$ near $x$. We can take this as reason to believe that the space part of the wave equation should be $\nabla^2 f$.

What should the time part be? I think it is rather clear that we need two time derivatives. For consider the concrete example of a mechanical wave where the motion is ultimately described by Newton's laws, which have two time derivatives. You would also expect that the time evolution of the wave would depend not only on its shape at $t =0$, but also on the time derivative at $t = 0$. Thus we the wave guess that waves are described by $$\frac{\partial^2}{\partial t^2}f - \frac{1}{c^2} \nabla^2 f = 0$$ where the quantity $c$ has dimensions of velocity, so as to make the equation dimensionally correct.


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